Double spring question

1. The problem statement, all variables and given/known data
A block of mass 20kg with a spring with a stiffness of 500N/m attached to the bottom is dropped .5m from rest onto another spring of stiffness 800N/m. The weight of the springs can be neglected. What is the maximum deformation of each spring due to the collision.

Ermm, if the springs are inline with each other (on top of each other) you add the reciprocals of the constants, so that:

[tex]{k_T}^{-1} = {k_1}^{-1} + {k_2}^{-1}[/tex]

The system adds (becomes stiffer) when the springs are used adjacent to each other.

So I think, from the wording of the question, you need to use this equation maybe:

[tex]k_T = \frac{k_1 k_2}{k_1 + k_2}[/tex] - Your thoughts

But yeah - the kinetic energy term would become [itex]mg\Delta h[/itex] in your equation (because the kinetic energy came about from falling the distance [itex]\Delta h[/itex], cause when it hits the spring, your [itex]mgh = 0[/itex]. Apart from that, logic seems fine...

I do agree with Sam.. Keq = k.k'/(k + k') , is the one to be used.
Since, the two springs are adjacent, force in them must be equal => k.x = k'.x'
Equivalent spring, Keq.X = k.x = k'.x', where X = x + x'.

@ sumeer@dinsum.c
I posted this, just to urge one thing: please REASON OUT your assumptions! Do not make blind assumptions. The assumptions you made (that Keq = k+ k') would have been true, if elongations in both the springs would have been same! x = x', X = x = x', and Ftot = Keq.X = k.x + k'.x'
I hope you get the idea.

Hi Sameer, I don't how you solved the problem. But see whether it matches with this. Loss of PE of block = mg(0.5 +X). This energy is shared by the two springs and it is equal to 1/2*kx^2 + 1/2*k'x'^2. During the compression force in each spring is kx and k'x'. Hence down ward force =mg + kx and upward force = k'x'. In equilibrium position mg +kx = k'x' and X = x+x', write x' and X in terms of x and substitute it in the above energy equation. Then solve for x, and x'